secx + log cos^2x .. max and min value ?
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f(x)=secx+log cosx²
f (x) = secx + 2 log cosx
Therefore, f′ (x) = secx tanx – 2 tanx = tanx (secx –2)
f ′ (x) = 0 ⇒ tanx = 0 or secx = 2 or cosx = 1/ 2
Therefore, possible values of x are x = 0, or x = π and x = π/3 or x = 5π / 3
f (x) = secx + 2 log cosx
Therefore, f′ (x) = secx tanx – 2 tanx = tanx (secx –2)
f ′ (x) = 0 ⇒ tanx = 0 or secx = 2 or cosx = 1/ 2
Therefore, possible values of x are x = 0, or x = π and x = π/3 or x = 5π / 3
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